JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Kinematics

A rubber ball is released from a height of 5 m above the floor. It bounces back repeatedly, always rising to

\frac{81}{100}

of the height through which it falls. Find the average speed of the ball. (Take

g = 10 ms^{- 2}

)

Answer: (B)

Solution

Each rebound reaches

\frac{81}{100}

of the height it fell through, so the heights form a geometric sequence

h_{n} = 5 (\frac{81}{100})^{n}

metres and the times, which go as

h

, fall by

\frac{9}{10}

each bounce. The total path length is the first drop plus a rise and a fall for every bounce:

D = h_{0} + 2 n \geq 1 \sum h_{n} = 5 + 2 (5) \frac{0.81}{1 - 0.81} = 5 (1 + \frac{162}{19}) = \frac{905}{19} m

The first drop takes

t_{0} = \frac{2 h _{0}}{g} = 1

s, and the

n

th up-and-down trip takes

2 t_{0} (\frac{9}{10})^{n}

, so

T = 1 + 2 \frac{0.9}{1 - 0.9} = 19 s

The average speed is

\frac{905/19}{19} = \frac{905}{361} = 2.50 m s^{- 1}

JEE Main · 2021 · 17 Mar 2021 · Shift 2 · Single correct option